Question

A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.09 hours, with a standard deviation of 2.29 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.31 hours, with a standard deviation of 1.65 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children (141 -H2) Let Hy represent the mean leisure hours of adults with no children under the age of 18 and H2 represent the mean leisure hours of adults with children under the age of 18. The 95% confidence interval for ( 14 - H2) is the range from hours to hours. (Round to two decimal places as needed.) What is the interpretation of this confidence interval? O A. There is 95% confidence that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours. B. There is a 95% probability that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours. OC. There is 95% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours. OD. There is a 95% probability that the difference of the means is in the interval. Conclude that there is a significant difference in the number of leisure hours.

Answer 1

1)

	Pop 1	Pop 2
sample mean x =	5.090	4.310
std deviation s=	2.290	1.650
sample size n=	40	40
Point estimate =x1-x2=		0.780
std error =√(S21/n1+S22/n2)=		0.4463

Point estimate of differnce =x1-x2     =		0.780
for 95 % CI & 39 df value of t=		2.023	from excel: t.inv(0.975,39)
margin of error E=t*std error                   =		0.903
lower bound=mean difference-E     =		-0.1228
Upper bound=mean differnce +E      =		1.6828
from above 95% confidence interval for population mean =(-0.12,1.68)

C. There is 95% confidence that the difference of the means is in the interval. Conclude that there is insufficient evidence of a significant difference in the number of leisure hours.

2)

because npo(1-po) =23.5 is greater than 10 , the sample size is less than or equal to 5%, of the population size, and the sample can be reasonably assumed to be random , the requirements for testing the hypothesis is satisfied.

null Hypothesis:              Ho:   p=		0.027
alternate Hypothesis:    Ha: p >		0.027

sample success x   =		28
sample size          n    =		894
std error σp =√(p*(1-p)/n) =		0.0054
sample prop p̂ = x/n=28/894=		0.0313
z =(p̂-p)/σp=(0.031-0.027)/0.005=		0.80
p value                          =		0.212	(from excel:1*normsdist(-0.8)

C. Since P-value > a, do not reject the null hypothesis and conclude that there is not sufficient evidence that more than 2.7% of the users experience flulike symptoms.